Math  /  Calculus

QuestionUse derivatives to find the critical points and inflection points. f(x)=5x6lnxf(x)=5 x-6 \ln x
Enter the exact answers in increasing order. If there is only one critical point, enter NA in the second area. If there are no inflection points, enter NA.
Critical points: x=x=\begin{array}{l} x=\square \\ x=\square \end{array}
Inflection point: x=x= \square

Studdy Solution

STEP 1

1. A critical point occurs where the first derivative is zero or undefined.
2. An inflection point occurs where the second derivative changes sign.
3. The natural logarithm function, lnx\ln x, is only defined for x>0x > 0.

STEP 2

1. Find the first derivative of f(x)f(x).
2. Determine the critical points by solving f(x)=0f'(x) = 0.
3. Find the second derivative of f(x)f(x).
4. Determine the inflection points by solving f(x)=0f''(x) = 0 and checking for sign changes.

STEP 3

Find the first derivative of f(x)=5x6lnxf(x) = 5x - 6 \ln x.
f(x)=ddx(5x)ddx(6lnx) f'(x) = \frac{d}{dx}(5x) - \frac{d}{dx}(6 \ln x)
f(x)=56x f'(x) = 5 - \frac{6}{x}

STEP 4

Determine the critical points by setting the first derivative equal to zero and solving for xx.
56x=0 5 - \frac{6}{x} = 0
5=6x 5 = \frac{6}{x}
x=65 x = \frac{6}{5}
Since x>0x > 0 for lnx\ln x to be defined, this is valid.

STEP 5

Find the second derivative of f(x)f(x).
f(x)=ddx(56x) f''(x) = \frac{d}{dx}\left(5 - \frac{6}{x}\right)
f(x)=0+6x2 f''(x) = 0 + \frac{6}{x^2}
f(x)=6x2 f''(x) = \frac{6}{x^2}

STEP 6

Determine the inflection points by setting the second derivative equal to zero and checking for sign changes.
6x2=0 \frac{6}{x^2} = 0
This equation has no solutions because 6x2\frac{6}{x^2} is never zero for any real xx. Therefore, there are no inflection points.
The critical point is at x=65x = \frac{6}{5}.
Critical points: x=65x=NA \begin{array}{l} x=\frac{6}{5} \\ x=\text{NA} \end{array}
There are no inflection points.
Inflection point: x=NA x = \text{NA}

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